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Question Number 120636 by TANMAY PANACEA last updated on 01/Nov/20

selective Binomial

$${selective}\:{Binomial} \\ $$

Commented by TANMAY PANACEA last updated on 01/Nov/20

Commented by TANMAY PANACEA last updated on 01/Nov/20

Commented by TANMAY PANACEA last updated on 01/Nov/20

Commented by TANMAY PANACEA last updated on 01/Nov/20

Commented by TANMAY PANACEA last updated on 01/Nov/20

Commented by TANMAY PANACEA last updated on 01/Nov/20

Commented by TANMAY PANACEA last updated on 01/Nov/20

Commented by Dwaipayan Shikari last updated on 01/Nov/20

S=(1/(1!(n−1)!))+(1/(3!(n−3)!))+... S=(n/(n!))+((n(n−1)(n−2))/(3!n!))+((n(n−1)(n−2)(n−3)(n−4))/(5!n!))+... =(1/(n!))(n+((n(n−1)(n−2))/(3!))+((n(n−1)(n−2)(n−3)(n−4))/(5!))+...) =(2^(n−1) /(n!)) (1+x)^n =1+nx+((n(n−1))/(2!))x^2 +... (1−x)^n =1−nx+((n(n−1))/(2!))x^2 −.. (((1+x)^n −(1−x)^n )/2)=nx+((n(n−1)(n−2))/(3!))x^3 +... 2^(n−1) =(n+((n(n−1)(n−2))/(3!))+...) (x=1)

$${S}=\frac{\mathrm{1}}{\mathrm{1}!\left({n}−\mathrm{1}\right)!}+\frac{\mathrm{1}}{\mathrm{3}!\left({n}−\mathrm{3}\right)!}+... \\ $$$${S}=\frac{{n}}{{n}!}+\frac{{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)}{\mathrm{3}!{n}!}+\frac{{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)\left({n}−\mathrm{3}\right)\left({n}−\mathrm{4}\right)}{\mathrm{5}!{n}!}+... \\ $$$$=\frac{\mathrm{1}}{{n}!}\left({n}+\frac{{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)}{\mathrm{3}!}+\frac{{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)\left({n}−\mathrm{3}\right)\left({n}−\mathrm{4}\right)}{\mathrm{5}!}+...\right) \\ $$$$=\frac{\mathrm{2}^{{n}−\mathrm{1}} }{{n}!} \\ $$$$\left(\mathrm{1}+{x}\right)^{{n}} =\mathrm{1}+{nx}+\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}!}{x}^{\mathrm{2}} +... \\ $$$$\left(\mathrm{1}−{x}\right)^{{n}} =\mathrm{1}−{nx}+\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}!}{x}^{\mathrm{2}} −.. \\ $$$$\frac{\left(\mathrm{1}+{x}\right)^{{n}} −\left(\mathrm{1}−{x}\right)^{{n}} }{\mathrm{2}}={nx}+\frac{{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)}{\mathrm{3}!}{x}^{\mathrm{3}} +... \\ $$$$\mathrm{2}^{{n}−\mathrm{1}} =\left({n}+\frac{{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)}{\mathrm{3}!}+...\right)\:\:\:\left({x}=\mathrm{1}\right) \\ $$

Commented by Dwaipayan Shikari last updated on 01/Nov/20

(1+x)^n =C_0 +C_1 x+... ∫_0 ^1 (1+x)^n =∫_0 ^1 C_0 +C_1 x+.. (2^(n+1) /(n+1))=C_0 +(1/2)C_1 +(1/3)C_2 +...(1/(n+1))C_n

$$\left(\mathrm{1}+{x}\right)^{{n}} ={C}_{\mathrm{0}} +{C}_{\mathrm{1}} {x}+... \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}+{x}\right)^{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} {C}_{\mathrm{0}} +{C}_{\mathrm{1}} {x}+.. \\ $$$$\frac{\mathrm{2}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}={C}_{\mathrm{0}} +\frac{\mathrm{1}}{\mathrm{2}}{C}_{\mathrm{1}} +\frac{\mathrm{1}}{\mathrm{3}}{C}_{\mathrm{2}} +...\frac{\mathrm{1}}{{n}+\mathrm{1}}{C}_{{n}} \\ $$

Commented by Dwaipayan Shikari last updated on 01/Nov/20

C_0 +(C_0 +C_1 )+(C_0 +C_1 +C_2 )+... n(C_0 +C_1 +C_2 +...)−(C_1 +2C_2 +3C_3 +..) n.2^n −n.2^(n−1) =n2^(n−1) (2−1)=n.2^(n−1) [(1+x)^n =C_0 +C_1 x+C_3 x^2 +.. n(1+x)^(n−1) =C_1 +2C_3 x+.. n.2^(n−1) =C_1 +2C_3 +...]

$${C}_{\mathrm{0}} +\left({C}_{\mathrm{0}} +{C}_{\mathrm{1}} \right)+\left({C}_{\mathrm{0}} +{C}_{\mathrm{1}} +{C}_{\mathrm{2}} \right)+... \\ $$$${n}\left({C}_{\mathrm{0}} +{C}_{\mathrm{1}} +{C}_{\mathrm{2}} +...\right)−\left({C}_{\mathrm{1}} +\mathrm{2}{C}_{\mathrm{2}} +\mathrm{3}{C}_{\mathrm{3}} +..\right) \\ $$$${n}.\mathrm{2}^{{n}} −{n}.\mathrm{2}^{{n}−\mathrm{1}} ={n}\mathrm{2}^{{n}−\mathrm{1}} \left(\mathrm{2}−\mathrm{1}\right)={n}.\mathrm{2}^{{n}−\mathrm{1}} \\ $$$$\left[\left(\mathrm{1}+{x}\right)^{{n}} ={C}_{\mathrm{0}} +{C}_{\mathrm{1}} {x}+{C}_{\mathrm{3}} {x}^{\mathrm{2}} +..\right. \\ $$$${n}\left(\mathrm{1}+{x}\right)^{{n}−\mathrm{1}} ={C}_{\mathrm{1}} +\mathrm{2}{C}_{\mathrm{3}} {x}+.. \\ $$$$\left.{n}.\mathrm{2}^{{n}−\mathrm{1}} ={C}_{\mathrm{1}} +\mathrm{2}{C}_{\mathrm{3}} +...\right] \\ $$

Commented by Dwaipayan Shikari last updated on 01/Nov/20

(1+x+2x^2 )^(20) =a_0 +a_1 x+... 1^(20) =a_0 (x=0) (2)^(40) =1+a_1 +a_2 +.... (x=1) 2^(20) =1−a_1 +a_2 −a_3 +....(x=−1) 2^(40) +2^(20) =2(1+a_2 +a_4 +...+a_(38) ) 2^(39) +2^(19) =a_0 +a_2 +.... 2^(19) (2^(20) +1)=a_0 +a_2 +...+a_(38)

$$\left(\mathrm{1}+{x}+\mathrm{2}{x}^{\mathrm{2}} \right)^{\mathrm{20}} ={a}_{\mathrm{0}} +{a}_{\mathrm{1}} {x}+... \\ $$$$\mathrm{1}^{\mathrm{20}} ={a}_{\mathrm{0}} \:\:\:\left({x}=\mathrm{0}\right) \\ $$$$\left(\mathrm{2}\right)^{\mathrm{40}} =\mathrm{1}+{a}_{\mathrm{1}} +{a}_{\mathrm{2}} +....\:\:\:\left({x}=\mathrm{1}\right) \\ $$$$\mathrm{2}^{\mathrm{20}} =\mathrm{1}−{a}_{\mathrm{1}} +{a}_{\mathrm{2}} −{a}_{\mathrm{3}} +....\left({x}=−\mathrm{1}\right) \\ $$$$\mathrm{2}^{\mathrm{40}} +\mathrm{2}^{\mathrm{20}} =\mathrm{2}\left(\mathrm{1}+{a}_{\mathrm{2}} +{a}_{\mathrm{4}} +...+{a}_{\mathrm{38}} \right) \\ $$$$\mathrm{2}^{\mathrm{39}} +\mathrm{2}^{\mathrm{19}} ={a}_{\mathrm{0}} +{a}_{\mathrm{2}} +.... \\ $$$$\mathrm{2}^{\mathrm{19}} \left(\mathrm{2}^{\mathrm{20}} +\mathrm{1}\right)={a}_{\mathrm{0}} +{a}_{\mathrm{2}} +...+{a}_{\mathrm{38}} \\ $$

Commented by TANMAY PANACEA last updated on 01/Nov/20

thank you

$${thank}\:{you} \\ $$

Commented by TANMAY PANACEA last updated on 01/Nov/20

i have posted questions collected from[books of IIT

$${i}\:{have}\:{posted}\:{questions}\:{collected}\:{from}\left[{books}\:{of}\:{IIT}\right. \\ $$

Commented by Dwaipayan Shikari last updated on 01/Nov/20

I am in class 11 (high school) :)

$$\left.{I}\:{am}\:{in}\:{class}\:\mathrm{11}\:\left({high}\:{school}\right)\::\right) \\ $$

Commented by prakash jain last updated on 02/Nov/20

You are doing very well for class XI. Are you aiming for IIT?

$$\mathrm{You}\:\mathrm{are}\:\mathrm{doing}\:\mathrm{very}\:\mathrm{well}\:\mathrm{for}\:\mathrm{class}\:\mathrm{XI}. \\ $$$$\mathrm{Are}\:\mathrm{you}\:\mathrm{aiming}\:\mathrm{for}\:\mathrm{IIT}? \\ $$

Commented by Dwaipayan Shikari last updated on 02/Nov/20

I want to study physics in later. IIT′s are good but i want to study freely

$${I}\:{want}\:{to}\:{study}\:{physics}\:{in}\:{later}. \\ $$$${IIT}'{s}\:{are}\:{good}\:{but}\:{i}\:{want}\:{to}\:{study}\:{freely} \\ $$

Commented by Dwaipayan Shikari last updated on 02/Nov/20

But in2022 i have to take this exam :)

$$\left.{But}\:{in}\mathrm{2022}\:{i}\:{have}\:\:{to}\:{take}\:{this}\:{exam}\::\right) \\ $$

Commented by prakash jain last updated on 02/Nov/20

Are u interested in theoritical physics?

$$\mathrm{Are}\:\mathrm{u}\:\mathrm{interested}\:\mathrm{in}\:\mathrm{theoritical}\:\mathrm{physics}? \\ $$

Commented by TANMAY PANACEA last updated on 02/Nov/20

17^(1983) +11^(1983) −7^(1983) answer is =1 explanation 7^1 =last digit 7 7^2 =l.d 9 7^3 =l.d 3 7^4 =1 7^5 =7 cycle of (7 9 1 7) ((1983)/7)=283+(2/7)→last digit=9 last digit of 7^(1983) →=9 last digit of 11^(1983) is1 so (9+1−9)→last digit=1 pls chk

$$\mathrm{17}^{\mathrm{1983}} +\mathrm{11}^{\mathrm{1983}} −\mathrm{7}^{\mathrm{1983}} \\ $$$${answer}\:{is}\:=\mathrm{1} \\ $$$${explanation} \\ $$$$\mathrm{7}^{\mathrm{1}} ={last}\:{digit}\:\mathrm{7} \\ $$$$\mathrm{7}^{\mathrm{2}} ={l}.{d}\:\mathrm{9} \\ $$$$\mathrm{7}^{\mathrm{3}} ={l}.{d}\:\mathrm{3} \\ $$$$\mathrm{7}^{\mathrm{4}} =\mathrm{1} \\ $$$$\mathrm{7}^{\mathrm{5}} =\mathrm{7} \\ $$$${cycle}\:{of}\:\left(\mathrm{7}\:\mathrm{9}\:\mathrm{1}\:\mathrm{7}\right) \\ $$$$\frac{\mathrm{1983}}{\mathrm{7}}=\mathrm{283}+\frac{\mathrm{2}}{\mathrm{7}}\rightarrow{last}\:{digit}=\mathrm{9} \\ $$$${last}\:{digit}\:{of}\:\mathrm{7}^{\mathrm{1983}} \rightarrow=\mathrm{9} \\ $$$${last}\:{digit}\:{of}\:\mathrm{11}^{\mathrm{1983}} \:{is}\mathrm{1} \\ $$$${so}\:\left(\mathrm{9}+\mathrm{1}−\mathrm{9}\right)\rightarrow{last}\:{digit}=\mathrm{1} \\ $$$$\boldsymbol{{pls}}\:\boldsymbol{{chk}} \\ $$

Commented by TANMAY PANACEA last updated on 02/Nov/20

32^(32^(32) ) devided by 7 (4×7+5)^(32^(32) ) =f(7)+5^(32^(32) ) now 32^(32) =even number 32^(32) 2^1 →2 2^2 →4 2^3 →8 2^4 →16 2^5 →32 cycle(last digit 2 4 8 6 ) 32^(32) =2k is a even number whose last digit 6 5^(2k) =last two digit is 25 so when devided by 7→last digit is 4 pls chk is my thinking correct

$$\mathrm{32}^{\mathrm{32}^{\mathrm{32}} } \:{devided}\:{by}\:\mathrm{7} \\ $$$$\left(\mathrm{4}×\mathrm{7}+\mathrm{5}\right)^{\mathrm{32}^{\mathrm{32}} } \\ $$$$={f}\left(\mathrm{7}\right)+\mathrm{5}^{\mathrm{32}^{\mathrm{32}} } \\ $$$${now}\:\mathrm{32}^{\mathrm{32}} ={even}\:{number} \\ $$$$\mathrm{32}^{\mathrm{32}} \\ $$$$\mathrm{2}^{\mathrm{1}} \rightarrow\mathrm{2} \\ $$$$\mathrm{2}^{\mathrm{2}} \rightarrow\mathrm{4} \\ $$$$\mathrm{2}^{\mathrm{3}} \rightarrow\mathrm{8} \\ $$$$\mathrm{2}^{\mathrm{4}} \rightarrow\mathrm{16} \\ $$$$\mathrm{2}^{\mathrm{5}} \rightarrow\mathrm{32} \\ $$$${cycle}\left({last}\:{digit}\:\:\:\mathrm{2}\:\mathrm{4}\:\mathrm{8}\:\mathrm{6}\:\right) \\ $$$$\mathrm{32}^{\mathrm{32}} =\mathrm{2}{k}\:{is}\:{a}\:{even}\:{number}\:{whose}\:{last}\:{digit}\:\mathrm{6} \\ $$$$\mathrm{5}^{\mathrm{2}{k}} ={last}\:{two}\:{digit}\:{is}\:\mathrm{25} \\ $$$${so}\:{when}\:{devided}\:{by}\:\mathrm{7}\rightarrow{last}\:{digit}\:{is}\:\mathrm{4} \\ $$$$\boldsymbol{{pls}}\:\boldsymbol{{chk}}\:\boldsymbol{{is}}\:\boldsymbol{{my}}\:\boldsymbol{{thinking}}\:\boldsymbol{{correct}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by Dwaipayan Shikari last updated on 02/Nov/20

yes , Prakash sir!

$${yes}\:,\:{Prakash}\:{sir}! \\ $$

Commented by Dwaipayan Shikari last updated on 03/Nov/20

(4×7+4)=32^(32^(32) ) typo

$$\left(\mathrm{4}×\mathrm{7}+\mathrm{4}\right)=\mathrm{32}^{\mathrm{32}^{\mathrm{32}} } \\ $$$${typo} \\ $$

Commented by TANMAY PANACEA last updated on 03/Nov/20

yes yes

$${yes}\:{yes}\: \\ $$

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